Question

The perimeter of a rectangle is 154m. Length is 2m more than twice its breadth. Find the length and breadth ?

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Answer 1

length =77m

breath =75m

Answer 2

Given : The perimeter of a rectangle is 154m & Length is 2m more than twice its breadth.

To Find : Find the length and breadth rectangle ?

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Solution : Let the breadth be x then its length be x + 2.

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$\underline{\frak{As ~we ~know ~that~:}}$

• $\boxed{\sf\pink{Perimeter~ of ~rectangle~=~2\bigg(l~+~b\bigg)}}$

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Where,

• "l" is length
• "b" is breadth

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❍ Perimeter of rectangle = 154m

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$\pmb{\sf{\underline{According ~to ~the~ Given ~Question~:}}}$

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$\qquad\qquad{\sf:\implies{2\bigg(x~+~2~x\bigg)~=~154}}$

$\qquad\qquad{\sf:\implies{2\bigg(2x~+~2\bigg)~=~154}}$

$\qquad\qquad{\sf:\implies{2x~+~2~=~\dfrac{154}{2}}}$

$\qquad\qquad{\sf:\implies{2x~+~2~=~\cancel\dfrac{154}{2}}}$

$\qquad\qquad{\sf:\implies{2x~+~2~=~77}}$

$\qquad\qquad{\sf:\implies{2x~=~77~-~2}}$

$\qquad\qquad{\sf:\implies{2x~=~75}}$

$\qquad\qquad{\sf:\implies{x~=~\dfrac{75}{2}}}$

$\qquad\qquad{\sf:\implies{x~=~\cancel\dfrac{75}{2}}}$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\purple{x~=~37.5m}}}}}$★

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Therefore,

• ${\sf{x~=~37.5m}}$

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Hence,

• $\underset{\blue {\bf Required\ Answer}}{\underbrace{\boxed{\frak{\pink{\dfrac{Length ~of ~rectangle~=~x~+~5~=~42.5m}{Breadth ~of ~rectangle~=~x~=~37.5m}}}}}}$

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$\pmb{\frak{More ~Information~:}}$

• ${\sf\leadsto{Volume~ of~ cylinder~=~πr^2h}}$

• ${\sf\leadsto{T.S.A ~of ~cylinder~=~2πrh~+~2πr^2}}$

• ${\sf\leadsto{Volume~ of ~cone ~=~\dfrac{1}{3}πr^2h}}$

• ${\sf\leadsto{C.S.A ~of ~cone~=~πrl}}$

• ${\sf\leadsto{T.S.A ~of~ cone~=~πrl~+~πr^2}}$

• ${\sf\leadsto{Volume ~of ~cuboid~=~l~×~b~×~h}}$

• ${\sf\leadsto{C.S.A ~of ~cuboid~=~2\bigg(l~+~b\bigg)h}}$

• ${\sf\leadsto{T.S.A ~of ~cuboid~=~2\bigg(lb~+~bh~+~lh\bigg)}}$

• ${\sf\leadsto{C.S.A ~of ~cube~=~4a^2}}$

• ${\sf\leadsto{T.S.A ~of ~cube~=~6a^2}}$

• ${\sf\leadsto{Volume~ of~ cube~=~a^3}}$

• ${\sf\leadsto{Volume ~of~ sphere~=~\dfrac{4}{3}πr^3}}$

• ${\sf\leadsto{Surface ~area ~of ~sphere~=~4πr^2}}$

• ${\sf\leadsto{Volume~ of ~hemisphere~=~\dfrac{2}{3}πr^3}}$

• ${\sf\leadsto{C.S.A ~of ~hemisphere~=~2πr^2}}$

• ${\sf\leadsto{T.S.A ~of ~hemisphere~=~3πr^2}}$

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